Let PQ be a focal chord of the parabola y2 = 4x such that it subtends an angle of $\frac{\pi }{2}$ at the point (3, 0). Let the line segment PQ be also a focal chord of the ellipse E: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$ $a^2>b^2.$ If e is the eccentricity of the ellipse E, then the value of $\frac{1}{e^2}$ is equal to:

Correct Option - 2


Detailed Solution:

 $AP\perp BP$

M₁ M₂ = 1

$\frac{2t}{t^2-3}\times \frac{2/6}{3-\frac{1}{t^2}}=-1$

t = 1

So, A (1, 2) and B (1, 2) they must be end pts of focal chord.

Length of latus rectum $=\frac{2b^2}{a}$

$4=\frac{2b^2}{a}$

b2 = 2a and ae = 1

Eccentricity of ellipse (Horizontal)

b² = a² (1 – e²)

2a = a² (1 – e²)

2 = $\frac{1}{e}\left(1-e^2\right)$

e² + 2e – 1 = 0

$e=\frac{-2\pm \sqrt[]{4+4}}{2}$

$e=-1+\sqrt[]{2}$

now $\frac{1}{e^2}$$=$$3$$+$$2$

<p>&nbsp;<math><mrow><mi>A</mi><mi>P</mi><mo>⊥</mo><mi>B</mi><mi>P</mi></mrow></math></p><p>M₁ M₂ = 1</p><p><math><mrow><mfrac><mrow><mn>2</mn><mi>t</mi></mrow><mrow><msup><mrow><mi>t</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>−</mo><mn>3</mn></mrow></mfrac><mo>×</mo><mfrac><mrow><mn>2</mn><mo>/</mo><mn>6</mn></mrow><mrow><mn>3</mn><mo>−</mo><mfrac><mrow><mn>1</mn></mrow><mrow><msup><mrow><mi>t</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac></mrow></mfrac><mo>=</mo><mo>−</mo><mn>1</mn></mrow></math></p><p>t = 1</p><div class="photo-widget-full"><div class="figure"><picture><source srcset="" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1752751158phpfjIDax.jpeg" alt="" width="262" height="260" loading="lazy"></picture></div></div><p>So, A (1, 2) and B (1, 2) they must be end pts of focal chord.</p><p>Length of latus rectum&nbsp;<math><mrow><mo>=</mo><mfrac><mrow><mn>2</mn><msup><mrow><mi>b</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><mi>a</mi></mrow></mfrac></mrow></math></p><p><math><mrow><mn>4</mn><mo>=</mo><mfrac><mrow><mn>2</mn><msup><mrow><mi>b</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><mi>a</mi></mrow></mfrac></mrow></math></p><p>b2 = 2a and ae = 1</p><p>Eccentricity of ellipse (Horizontal)</p><p>b²= a² (1 – e²)</p><p>2a = a² (1 – e²)</p><p>2 =&nbsp;<math><mrow><mfrac><mrow><mn>1</mn></mrow><mrow><mi>e</mi></mrow></mfrac><mrow><mo> (</mo><mrow><mn>1</mn><mo>−</mo><msup><mrow><mi>e</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mo>)</mo></mrow></mrow></math></p><p>e²+ 2e – 1 = 0</p><p><math><mrow><mi>e</mi><mo>=</mo><mfrac><mrow><mo>−</mo><mn>2</mn><mo>±</mo><mroot><mrow><mn>4</mn><mo>+</mo><mn>4</mn></mrow><mrow></mrow></mroot></mrow><mrow><mn>2</mn></mrow></mfrac></mrow></math></p><p><math><mrow><mi>e</mi><mo>=</mo><mo>−</mo><mn>1</mn><mo>+</mo><mroot><mrow><mn>2</mn></mrow><mrow></mrow></mroot></mrow></math></p><p>now&nbsp;<math><mrow><mfrac><mrow><mn>1</mn></mrow><mrow><msup><mrow><mi>e</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac></mrow></math><math><mrow><mo>=</mo></mrow></math><math><mrow><mn>3</mn></mrow></math><math><mrow><mo>+</mo></mrow></math><math><mrow><mn>2</mn></mrow></math></p>

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